Number of domino tilings in K_3 X P_2n (or in S_4 X P_2n).
Number of perfect matchings in graph C_{3} X P_{2n}.
Number of perfect matchings in S_4 X P_2n.
In general, Sum_{k=0..n} binomial(2*n-k, k)*j^(n-k) = (-1)^n * U(2*n, i*sqrt(j)/2), i=sqrt(-1). -
Paul Barry, Mar 13 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4} which do not end in 0 (e.g., at n=2, we have 02, 03, 04, 11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44). -
Tanya Khovanova, Jan 10 2007
(sqrt(21)+5)/2 = 4.7912878... = exp(arccosh(5/2)) = 4 + 3/4 + 3/(4*19) + 3/(19*91) + 3/(91*436) + ... -
Gary W. Adamson, Dec 18 2007
a(n+1) is the number of compositions of n when there are 4 types of 1 and 3 types of other natural numbers. -
Milan Janjic, Aug 13 2010
For n >= 2, a(n) equals the permanent of the (2n-2) X (2n-2) tridiagonal matrix with sqrt(3)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. -
John M. Campbell, Jul 08 2011
Values of x (or y) in the solutions to x^2 - 5xy + y^2 + 3 = 0. -
Colin Barker, Feb 04 2014
All positive solutions of the Diophantine equation x^2 + y^2 - 5*x*y = -3 (see the preceding comment) are given by [x(n) = S(n, 5) - S(n-1, 5), y(n) = x(n-1)], for n =-oo..+oo, with the Chebyshev S-polynomials (
A049310), with S(-1, 0) = 0, and S(-|n|, x) = - S(|n|-2, x), for |n| >= 2.
This binary indefinite quadratic form has discriminant D = +21. There is only this family representing -3 properly with x and y positive, and there are no improper solutions.
See the formula for a(n) = x(n-1), for n >= 1, in terms of S-polynomials below.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in
A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)
All proper and improper solutions of the generalized Pell equation X^2 - 21*Y^2 = +4 are given, up to a combined sign change in X and Y, in terms of x(n) = a(n+1) from the preceding comment by X(n) = x(n) + x(n-1) = S(n-1, 5) - S(n-2, 5) and Y(n) = (x(n) - x(n-1))/3 = S(n-1, 5), for all integer numbers n. For positive integers X(n) =
A003501(n) and Y(n) =
A004254(n). X(-n) = X(n) and Y(-n) = - Y(n), for n >= 1.
The two conjugated proper families of solutions are given by [X(3*n+1), Y(3*n+1)] and [X(3*n+2), Y(3*n+2)], and the one improper family by [X(3*n), Y(3*n)], for all integer n. This follows from the mentioned paper by Robert K. Moniot. (End)
Equivalent definition: a(n) = ceiling(a(n-1)^2 / a(n-2)), with a(1)=1, a(2)=4, a(3)=19. The problem for USA Olympiad (see Andreescu and Gelca reference) asks to prove that a(n)-1 is always a multiple of 3. -
Bernard Schott, Apr 13 2022